## KingGeorge 2 years ago [SOLVED] Here's a cute predicate logic problem. Prove or disprove: $\forall x \,\exists y \,\forall z\, \exists w \;\;\text{such that} \;\;x+y+z+w=xyzw$Assume that $$x, y, z, w \in \mathbb{C}$$.

1. experimentX

is xyzw a combination of digit or multiplication of digit??

2. KingGeorge

multiplication

3. experimentX

lol .. didn't realize they were complex

4. KingGeorge

It doesn't really matter. I just put that specification on there so that know one tried to change the problem so that addition and multiplication had different meanings.

5. KingGeorge

Just prove/disprove it for $$\mathbb{R}$$ and it will become clear why it doesn't matter if they're complex numbers.

6. KingGeorge

Hint below in white LaTeX$\color{white}{\text{There is a certain value for y that you should choose to prove it.}}$

7. experimentX

$\color{white}{\text{LOL i didn't know that!!.}}$

8. KingGeorge

Another, probably more helpful hint, below in white LaTeX$\color{white}{\text{What value of y can you choose so that xyzw is constant?}}$

9. nbouscal

Choose y=0, then choose w=-(x+z).

10. KingGeorge

Bingo.

11. experimentX

But ∃y doesn't mean that y is rather a variable such that we have to fit constraint ??

12. KingGeorge

$$\exists\, y$$ means that we can choose a specific value for y. It's a variable, but one that we have complete control over.

13. experimentX

then it must be false ... i guess so!!

14. KingGeorge

The "there exists" only means that at least one possible value of y where this is true. One value that we know works is $$y=0$$. This means that there does exists at least one value for $$y$$ that makes the statement true, so we have proven that the original statement is true.

15. experimentX

does't y=0 make statement untrue?? xyzw=0, for y=0

16. nbouscal

Right, both the sum and the multiplication end up equaling zero because you set w=-(x+z).

17. KingGeorge

If $$y$$ is 0, and $$x=-(x+z)$$ we have $x+0+z+(-(x+z))=x+z-x-z=0$and $x\cdot0\cdot z\cdot w=0$So that means that $x+y+z+w=xyzw$

18. experimentX

Oh ... stupid me ... i never considered the negative values!!

19. KingGeorge

Hope everyone enjoyed this little (unintuitive) problem. Good night.

20. experimentX

thanks!! i'll try to think broader way next time!!

21. KingGeorge

I'll try to get a problem just for you next time then. Any preference about the subject?

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